如何在get（）調用中將條件傳遞給「包含」？

Question

當獲取單個記錄和相關記錄（下兩級）時，有什麼方法可以將條件傳遞給CakePHP 3中包含的記錄？在這種情況下，例如：

$user = $this->Users->get($this->Auth->user('id'), [ 
    'contain' => [ 
     'Articles' => ['Comments'] 
    ] 
]); 

如果嘗試了一堆有回調的方式，但似乎只在第一個層次上工作。像這樣，例如：

$user = $this->Users->get($this->Auth->user('id'), [ 
    'contain' => [ 
     'Articles' => function ($q) { 
      return $q->where(['Articles.published' => true]); 
     } 
    ] 
]); 

有什麼想法？

Answer 1

，你應該能夠做到：

$user = $this->Users->get($this->Auth->user('id'), [ 
    'contain' => [ 
     'Articles' => function ($q) { 
      return $q->where(['Articles.published' => true]); 
     }, 
     'Articles.Comments' => function ($q) { 
      return $q->where(['Comments.deleted' => false]); 
     } 
    ] 
]); 

或

$user = $this->Users->get($this->Auth->user('id'), [ 
    'contain' => [ 
     'Articles' => function ($q) { 
      $q->contain([ 
       'Comments' => function ($q) { 
        return $q->where(['Comments.deleted' => false]); 
       } 
      ]); 
      return $q->where(['Articles.published' => true]); 
     }    
    ] 
]); 

Answer 2

@ arilia的選項應該可以正常工作。我的選擇是

$user = $this->Users->get($this->Auth->user('id'), [ 
    'contain' => [ 
     'Articles' => [ 
      'queryBuilder' => function ($q) { 
       return $q->where(['Articles.published' => true]); 
      }, 
      'Comments' => [ 
       'querybuilder' => function ($q) { 
        return $q->where(['Comments.deleted' => false]); 
       } 
      ], 
     ], 
    ], 
]); 

（試過張貼此作爲一個評論，而不是一個單獨的答案，但它不會格式化代碼正確那裏。）